WHY can't I solve this! It's a super-simple question! Please help!?

Question

HI! This is so stupid that I can't stand it!!! I don't know why I can't solve it!

Okay, we have a ball that is dropped from a height of 294 feet that can drop at a rate of 34.3 feet/second. The height, s, can be calculated by the formula:

s(t) = 4.9t² + 34.3t

where t is the time in seconds. How long does it take for the ball to reach the ground?

I tried a few different ways (why doesn't the quadratic formula work?), but nothing comes out right! Please explain how you solve this (without graphing it). Okay thanks! I'll pick a best answer TODAY!

Okay, I just figured out what I did wrong. I hit X 2 on my calculator instead of X^2. That was so stupid! That's why I was getting complex numbers... So thanks but never mind! hahaha I figured it out!

Answer 1

4.9t^2+34.3t-294=0

the quadratic formula works just fine.

t= 5s

Answer 2

Setting aside the accuracy of your physics...

s(t) is 294 feet
s(t) = 4.9t^2 + 34.3t
294 = 4.9t^2 + 34.3t

4.9t^2 + 34.3t - 294 = 0

Your basic quadratic: Ax^2 + Bx = C = 0
Here, A = 4.9 and B = 34.3 and C = -294

Plug them in to

x = (-B +/- sqrt(B^2 - 4AC))/2A

You get x = 5 or x = -12

-12 seconds makes no physical sense, so five seconds is your answer

Answer 3

Well to start with you are mixing units. Your 4.9 multiplying t^2 is 9.8/2 m/s^2 yet you used speed in ft/s. TO be consistent, use g = 32 ft/s^2 so that

s(t) = 16t^2 + 34.3t = 294 ft

t = -34.3/2 +/- sqrt( (34.3)^2 +4*16*294)

Answer 4

s(t) = 4.9t² + 34.3t = 294 ft, the depth from the starting position
Solve 4.9t² + 34.3t - 294 = 0 for t,
t = 5 sec
------------
It's better to say s(t) is the depth.

Answer 5

you don't need the quadratic equation to calculate the time.

The time is 294/34.3 = 8.57 seconds.

Answer 6

H/R=T

Hight divided by Rate of speed equals Time.

294ft / 34.3ftps= 8.5714285714285714285714285714286 seconds of flight.

Answer 7

whoa...